C - Writing a swap function for bubblesort

[Travercraig]

Okay I've been given a bubblesort function and i've wrote a main program to use bubblesort. However I also have to write a swap function, I think i understand what to put inside the swap function body, but i'm not sure what paramters to put in the swap function. I'll show you the code. I've commented what needs to be done.

 

#include <stdio.h>
#include <stdlib.h>

void SWAP ()//I don't know what paramtaers to put in the swap function

void bubblesort (int a[], int N)
{
     int i, j;
     
     for (i = N - 1; i > 0; i--)
         for (j = 0; j < i; j++)
             if (!(a[j] <= a[j+1]))
                SWAP(a,j,j+1);//a swap function for here, I'm not sure how this part works        
}

int main()
{
	int array[], number, i, j;
	
	printf("Enter the number of elements: ");
	scanf("%i", &number);
	
	prinf("You can now enter %i elements: ", number)
	
    for (i = 0; i < number: i++;)
        scanf("%i", array[i]);
	
	bubblesort(array, nuumber);
	
	for (j  = 0; j < number; j++ )
        printf("%ld\n", array[c]);
 
    return 0;   
}

 

[Sauron]

It's just the function that swaps the two elements your sort selected, all you need to do is to save one of the elements in a buffer, copy the second element where the first was and then copy the buffer where the second was.

[Travercraig]

14 minutes ago, Sauron said:

It's just the function that swaps the two elements your sort selected, all you need to do is to save one of the elements in a buffer, copy the second element where the first was and then copy the buffer where the second was.

I know that part I'm just unsure how I should layout the parameters in the swap function declaration.

[Unimportant]

Generally, one provides pointers to the actual elements to swap to a swap function so you don't have to pass along the array itself.

 

void SWAP(int *a, int *b)
{
	int Temp = *a;	
  	*a = *b;
  	*b = Temp;
}

Then you can call it as such:

SWAP(a + j, a + j + 1);

Wich is a, the array, which decays into a pointer + j, the offset.

a pointer + an offset results in a new pointer to the object at position j in the array.

[Sauron]

16 minutes ago, GR412 said:

I know that part I'm just unsure how I should layout the parameters in the swap function declaration.

declare a buffer integer, set it to be equal to a[j], then make a[j] equal to a[j+1] make a[j+1] equal to the buffer.

[Travercraig]

19 minutes ago, Sauron said:

declare a buffer integer, set it to be equal to a[j], then make a[j] equal to a[j+1] make a[j+1] equal to the buffer.

Can you show me what you mean because i'm under the impression you've still giving me the swap function body code, not the declaration parameters.

[Travercraig]

31 minutes ago, Unimportant said:

Generally, one provides pointers to the actual elements to swap to a swap function so you don't have to pass along the array itself.

 

void SWAP(int *a, int *b)
{
	int Temp = *a;	
  	*a = *b;
  	*b = Temp;
}

Then you can call it as such:

SWAP(a + j, a + j + 1);

Wich is a, the array, which decays into a pointer + j, the offset.

a pointer + an offset results in a new pointer to the object at position j in the array.

I'm trying to do it with the code i've been given, ther swap call for it is: SWAP(a,j,j+1); I need to somhow implement that call into a swap function.

[mathijs727]

void swap(auto* pBuffer, int i, int j)
{
    auto tmp = pBuffer;
    pBuffer = pBuffer[j];
    pBuffer[j] = tmp;
}

 

[Sauron]

33 minutes ago, GR412 said:

Can you show me what you mean because i'm under the impression you've still giving me the swap function body code, not the declaration parameters.

Oh wait, you mean whether you need a pointer or something else? Arrays can be passed as a copy of the array (which is what you DON'T want) or as pointers. You can find more precise information and examples here https://www.tutorialspoint.com/cprogramming/c_passing_arrays_to_functions.htm

 

j and j+1 are just integers and there's no harm in passing them as copies of the original, so you can write that the second and third parameters are just ints.

[Unimportant]

2 hours ago, mathijs727 said:

void swap(auto* pBuffer, int i, int j)
{
    auto tmp = pBuffer;
    pBuffer = pBuffer[j];
    pBuffer[j] = tmp;
}

 

afaik, "auto" is C++ (11 or higher) only. For C it should be "int tmp".

[Travercraig]

2 hours ago, Unimportant said:

afaik, "auto" is C++ (11 or higher) only. For C it should be "int tmp".

Ahh no wonder i got confused, so it would be like this: void swap (int tmp, int i, int j)?

 

[Nineshadow]

void swap(int *v, int i, int j)
{
	int tmp=v[i];
	v[i]=v[j];
	v[j]=tmp;
}

This will swap the elements at the i and j index of an array v, which seems to be what you want to do.